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vesna_86 [32]
3 years ago
6

Which of the following are roots of the polynomial function? Check all that apply f(x) = x^3 + 3x^2 - 9x + 5

Mathematics
1 answer:
lakkis [162]3 years ago
8 0

Answer:

x = 1 or x = -5

Step-by-step explanation:

Solve for x:

x^3 + 3 x^2 - 9 x + 5 = 0

The left hand side factors into a product with two terms:

(x - 1)^2 (x + 5) = 0

Split into two equations:

(x - 1)^2 = 0 or x + 5 = 0

Take the square root of both sides:

x - 1 = 0 or x + 5 = 0

Add 1 to both sides:

x = 1 or x + 5 = 0

Subtract 5 from both sides:

Answer:  x = 1 or x = -5

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Considering a discrete distribution, it is found that the expected number of new employees to be hired by the​ airline is of 446.16.

<h3>What is the mean of a discrete distribution?</h3>

The expected value of a discrete distribution is given by the <u>sum of each outcome multiplied by it's respective probability</u>.

In this problem, considering the situation described in the text, the distribution is given by:

  • P(X = 837) = 0.37.
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Hence, the expected value is given by:

E(X) = 0.37(867) + 0.63(199) = 446.16.

More can be learned about discrete distributions at brainly.com/question/24855677

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2 years ago
Which equivalent fractions can you use to find the sum of 2/8 + 3/9?
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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
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Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

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azamat
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 Round to the previous whole number.
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they can make about 7 teams.
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4 years ago
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